A) \[1-2({{2}^{n}})\]
B) \[{{2}^{n+3}}-\frac{1}{4}\]
C) \[\frac{1}{{{2}^{n+1}}}\]
D) \[\frac{1}{{{2}^{n-1}}}\]
Correct Answer: C
Solution :
[c] \[\frac{{{2}^{(n+2)}}-2\left( {{2}^{n}} \right)}{{{2}^{(2n+2)}}}=\frac{{{2}^{n}}\cdot {{2}^{2}}-{{2.2}^{n}}}{{{2}^{2}}\cdot {{2}^{2n}}}=\frac{{{2.2}^{n}}\left( 2-1 \right)}{{{2}^{2}}{{.2}^{2n}}}\] \[=\frac{1}{{{2.2}^{n}}}=\frac{1}{{{2}^{(n+1)}}}\]You need to login to perform this action.
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